The half-life of a radioactive nucleus is 50 | vedclass

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(D) Let $N_0$ be the initial number of radioactive nuclei.At time $t_1$,the amount decayed is $\frac{1}{3}N_0$,so the remaining amount is $N(t_1) = N_

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$50$

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(D) Let $N_0$ be the initial number of radioactive nuclei.At time $t_1$,the amount decayed is $\frac{1}{3}N_0$,so the remaining amount is $N(t_1) = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0$.At time $t_2$,the amount decayed is $\frac{2}{3}N_0$,so the remaining amount is $N(t_2) = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0$.We observe that $N(t_2) = \frac{1}{2} N(t_1)$.Since the number of active nuclei is halved in the interval $(t_2 - t_1)$,this time interval must be equal to the half-life $T_{1/2}$.Therefore,$t_2 - t_1 = T_{1/2} = 50 \, 	ext{days}$.

The half-life of a radioactive nucleus is $50$ days. The time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{2}{3}$ of it had decayed and the time $t_1$ when $\frac{1}{3}$ of it had decayed is (in days):

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